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I have some questions on scheme morphisms. I ask pardon for posting them in one thread as they are most likely not worth to be distributed into several threads.

Let $X=Spec R$ be a noetherian scheme.

  1. For a maximal ideal $m$ of $R$ one has a morphism $Spec (k(m))\to X$ induced by the ring homomorphism $R\to R_m\to R_m/R_mm=R/m$ which is a closed immersion. It maps the only point of $Spec (k(m))$ to $m$. If one takes a prime ideal $p$ instead of a maximal ideal, there is a scheme morphism induced by $R\to R_p\to R_p/R_p p=Quot(R/p)$. Is the image of the only point $(0)$ of $Spec (Quot(R/p))$ the generic point of the irreducible subscheme of $X$ correspoinding to $p$?

  2. For a prime ideal $p$ one has, as above, scheme morphisms $$ Spec(Quot(R/p))\xrightarrow{g_p} Spec(R_p)\xrightarrow{f_p} X $$ and if $p$ is maximal $Spec(Quot(R/p))$ can be thought of as a point. What is $Spec(R_p)$ geometrically? Is $f_p$ or $g_p$ respectively an open/closed immersion? How can the image of $f_p$ or $g_p$ respectively be thought of?

  3. If one has a morphism $f:Y\to X$, the base change of $f$ by $Spec (k(m))\to X$ is the fiber of $f$ at the point $m$. Is the base change of $f$ by $Spec(R_m)\to X$ a kind of 'thickened fiber'? In how far can I reconstruct the morphism $f$ if I know what it does on the topological spaces and if I know all the fibers or 'thickened fibers' respectively?

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1. Yes, because this image is $p$, which is indeed the generic point of $V(p)$, the subscheme of $X$ corresponding to $p$.

2. The affine scheme $Spec (R_p) $ is exactly the intersection of all the neighbourhoods of $p$ in $X$. It is neither closed nor open in general. It is an incarnation of the germ of the scheme $X$ at $p$. So $f_p$ is neither a closed nor an open immersion but some sort of generalized immersion.

The morphism $g_p$ is the closed immersion of the closed point $pR_p$ of the local ring $R_p$ into the spectrum $Spec(R_p)$ of said local ring.

3. Yes, your intuition is excellent:
for every $p\in X=Spec(R)$ the scheme $Y_p=f^{-1} (Spec(R_p)=Spec(R_p)\times_X Y$ is indeed a thickening of the fibre $Y(p)=f^{-1}(p)=Spec(\kappa(p))\times_X Y$ of $p$ , which has the following satisfying property.
Given $y\in Y$ over $p$, the local ring $\mathcal O_{Y_p,y}$ of this thickened fibre is exactly the same as the original local ring $\mathcal O_{Y,y}$ , in contradistinction to the local ring of the fibre $Y(p)$ at $y$ which is $\mathcal O_{Y(p),y}=\mathcal O_{Y,y}/pO_{Y,y}$ . In other words, the thickened fibre behaves as if it were the inverse image of some open neighbourhood of $p$, although it is actually smaller than any such inverse image.
Hence indeed you can reconstruct the morphism if you know all these thickenings (but beware, I haven't written down the details).

On the whole, I must congratulate you on the lucidity and pertinence of your questions: they are remarkable for a geometrystudent !

Edit: Ah ha, I have found a reference in the Holy Text for the fact that a morphism is indeed determined by its restrictions to the thickenings: EGA, I, Proposition 6.5.1.(i)

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